\newproblem{lay:2_1_3}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.1.3}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Let $A=\begin{pmatrix}2 & -5 \\ 3 & -2\end{pmatrix}$. Calculate $3I_2-A$ and $(3I_2)A$
}{
   % Solution
	\begin{center}
		$\begin{array}{rcl}
		   3I_2-A&=&3\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}-\begin{pmatrix}2 & -5 \\ 3 & -2\end{pmatrix}=
			          \begin{pmatrix}3 & 0 \\ 0 & 3\end{pmatrix}-\begin{pmatrix}2 & -5 \\ 3 & -2\end{pmatrix}=\begin{pmatrix}1 & 5 \\ -3 & 5\end{pmatrix} \\
		   (3I_2)A&=&\left(3\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}\right)\begin{pmatrix}2 & -5 \\ 3 & -2\end{pmatrix}=
			          \begin{pmatrix}3 & 0 \\ 0 & 3\end{pmatrix}\begin{pmatrix}2 & -5 \\ 3 & -2\end{pmatrix}=\begin{pmatrix}6 & -15 \\ 9 & -6\end{pmatrix} \\
		\end{array}$
	\end{center}
}
\useproblem{lay:2_1_3}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
